Question

Partial fractions?

how does dz/[z(1-vz)] becomes (dz/z) + (vdz)/(1-vz) using partial fractions? Strangely I only get dz/(1-vz) after partial fractions..

Answer 1

\[\frac{ dz }{ z(1-vz) }=(\frac{ 1 }{ z(1-vz) })dz\]\[(\frac{ A }{ z }+\frac{ B }{ 1-vz })dz=(\frac{ 1 }{ z(1-vz) })dz\]\[(\frac{A(1-vz)+ B(z) }{ (1-vz )})dz=(\frac{ 1 }{ z(1-vz) })dz\]\[(\frac{A-Avz+ Bz }{z (1-vz )})dz=(\frac{ 1 }{ z(1-vz) })dz\]\[(\frac{A-(Av- B)z }{z (1-vz )})dz=(\frac{ 1 }{ z(1-vz) })dz\]Now what U must notice here is that the right and left hand side expressions must be similar such that
A = 1 and Av-B = 0
Thus A=1 & B= v
Finally insert the values of A and B in the second step of the calculation.

Answer 2

Oh I see. Thanks so much @Zekarias!

Answer 3

Btw, I was wondering how you would integrate the equation (dz/z) + (vdz)/(1-vz)? I am confused because of the derivatives..

Answer 4

dz/z is the same thing as 1/z dz
You don't remember that one? :)
Important integral to remember, it comes up over and over again.

Answer 5

Inz, I believe? So it would become zIn(z), but I don't know how to integrate vdz/(1-vz)

Answer 6

v is a constant?

Answer 7

ν be the rate at which people who contract smallpox die from the disease (it's part of the question)

Answer 8

\[\int\limits_{}^{}(1/x) dx=\ln x\]
The other one is just another natural log, :D don't let the extra letters confuse you.
\[\int\limits_{}^{}\frac{ vdz }{ 1-vz }=-v \ln(1-vz)\]
Doing a substitution might help you to get through it :)

Answer 9

Grr screwed that one up :O lol i shoulda done a substitution XD

Answer 10

nope it is -ln (1-vz)

Answer 11

Do you integrate it like how you would integrate an equation as if the v was an constant? Because v is a rate, it's kind of confusing

Answer 12

What did z imply?

Answer 13

It's a constant rate. Otherwise the problem doesn't really make sense. :D Since you don't have any dv's

Answer 14

Oh, I got it. I did u=1-vz and du=-vdz which would be vdz=-du. So altogether it became -In(1-vz)